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4 years ago

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A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, start

ing from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?

1 answer:

Len [333]4 years ago

8 0

Answer:

a) 16.46 seconds.

b) 13.46 seconds

c) 2.67 m/s²

Explanation:

Acceleration = a = 1.35 m/s²

Final velocity = v = 80 km/h = $80\frac{1000}{3600}=22.22\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 22.22=0+1.35t\\\Rightarrow t=\frac{22.22}{1.35}=16.46\ s$

Time taken to accelerate to top speed is 16.46 seconds.

Acceleration = a = -1.65 m/s²

Initial velocity = u = 80 km/h= $80\frac{1000}{3600}=22.22\ m/s$

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=22.22-1.65t\\\Rightarrow t=\frac{22.22}{1.65}=13.46\ s$

Time taken to stop the train from top speed is 13.46 seconds

Initial velocity = u = 80 km/h= $80\frac{1000}{3600}=22.22\ m/s$

Time taken = t = 8.3 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=22.22+a8.3\\\Rightarrow a=\frac{-22.22}{8.3}=-2.67\ m/s^2$

Emergency deceleration is 2.67 m/s²

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As wee know that the amplitude of the wave will decide the energy of the wave

Here we know that energy density of electromagnetic wave is given as

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now we have

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so here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa.

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4 years ago

Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i

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Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = $N_{p}$ = 500 turns

input voltage = $V_{p}$ = 120 V

number of secondary turns = $N_{s}$ = 3 turns

output voltage = $V_{s}$ = ?

using the equation for a transformer

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

substituting values, we have

$\frac{V_{s} }{120 } = \frac{3 }{500} }$

$500V_{p} = 120*3\\500V_{p} = 360$

$V_{p}$ = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

$I_{P}V_{p} = I_{s}V_{s}$

where

$I_{p}$ = input current = ?

$I_{s}$ = output voltage = 3.2 A

$V_{s}$ = output voltage = 0.72 V

$V_{p}$ = input voltage = 120 V

substituting values, we have

120 $I_{p}$ = 3.2 x 0.72

120 $I_{p}$ = 2.304

$I_{p}$ = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = $I_{p} V_{p}$

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3 years ago

Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one

kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

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distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

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Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

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F₁ = 1.35 x 10⁻¹³ N

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Electric force on charge due to rod 2:

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3 years ago

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Answer:

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Explanation:

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In the proposed case the shock is when the thickness changes, in this case we have the above phenomena, a part of the wave is reflected by being inverted and a part of the wave is transmitted without inverting.

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When checking the answers the correct one is the reflected wave is inverted and the transmitted wave is up

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